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Let I_{3} be the clockwise loop current in center loop
V_{2} = 4(I_{2} + I_{3}) ⇒ I_{3} = 0.25V_{2}  I_{2}
⇒ I_{1} = 0.35V_{2}  I_{2} .........(i)
V_{1} = 4I_{1}  0.2V_{1} + V_{2}
1.2V_{1} = 4(0.35V_{2}  I_{2}) + V_{2} = 2.4V_{2}  4I_{2}
⇒ V_{1} = 2V_{2}  3.33I_{2} .........(ii)
..........(i)
I_{1} = (V_{1}  V_{2}) Y_{ab} + V_{1}Y_{ab}
⇒ I_{1} = V_{1}(Y_{a} + Y_{ab})  V_{2}Y_{ab} .....(i)
I_{2} = (V_{2}  V_{1})Y_{ab} + V_{2}Y_{b} = V_{1}Y_{ab} + V_{2}(Y_{b} + Yab) .....(ii)
The yparameters of a 2port network are
A resistor of 1 ohm is connected across as shown in fig. The new y –parameter would be
yparameter of 1Ω resistor network are
New yparameter
V_{1} = 600I_{1} + 100I_{2} , V_{2} = 100I_{1} + 200I_{2}
V_{s} = 60I_{1} + V_{1} = 660I_{1} + 100I_{2} , V_{2} = V_{o} = 300I_{2}
_{}
The Tparameters of a 2port network are
If such two 2port network are cascaded, the z –parameter for the cascaded network is
.......(i)
........(ii)
Which among the following is regarded as short circuit forward transfer admittance?
In the circuit shown in fig. P.1.10.34, when the voltage V_{1} is 10 V, the current I is 1 A. If the applied voltage at port2 is 100 V, the short circuit current flowing through at port 1 will be
Interchaning the port
For a 2port symmetrical bilateral network, iftransmission parameters A = 3 and B = 1Ω , the value ofparameter C is
For symmetrical network A = D = 3
For bilateral AD  BC = 1, 9  C = 1,C = 8 S
A network has 7 nodes and 5 independent loops. The number of branches in the network is
V_{1} = 0.5V_{1} + I_{1} + 2(I_{1} + I_{2}) + aI_{1}
⇒ V_{1} = (6 + 2a)I_{1} + 4I_{2} ......(i)
V2 = 2(I_{1} + I_{2}) + aI_{1} ⇒ V_{2} = (2 + a)I_{1} + 2I_{2} ......(ii)
For reciprocal network
z_{12} = z_{21}, 4 = 2 + a ⇒ a = 2
I_{1} = 4 x 10^{3}V_{1} 0.1 x 10^{3}V_{2}
I_{2} = 50 X 10^{3}V_{1} + 10^{3}V_{2} ,V_{2} = 10^{3}I_{2}
 10^{3}V2 = 50 x 10^{3}V_{1} + 10^{3}V_{2}, V_{2} = 25V
I_{1} = 10 x 10^{3}V_{1} 5 x 10^{3}V_{2},
100 = 25I_{1} + V_{1}
100  V_{1} = 0.25V_{1}  0.25V_{2} ⇒ 800 = 10V_{1}  V_{2 } ....(i)
I_{2} = 50 x 10^{3}V_{1} + 20 x 10^{3}V_{2}, V_{2} = 100I_{2}
V_{2} = 5V_{1} V_{2} ⇒ 3V_{2} + 5V_{1} = 0 ....(ii)
From (i) and (ii) V_{1} = 68.6 V, V2 = 114.3 V.
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